You must be more careful in deducing the likelihood function, otherwise your thinking seems correct. The density function for one $X_i$ is $$ f(x_i)=\frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} $$ So the likelihood function can be written as $$ \mathcal{L}(\theta)=\prod_{i=1}^n \frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} \\ =2^{-n}\theta^{-n} \cdot \prod_i I\{-\theta<X_i < \theta \} \\ = 2^{-n}\theta^{-n} \cdot I\{-\theta < \min_i x_i \le \max_i x_i <\theta \} \\ = 2^{-n}\theta^{-n} \cdot I\{0\le \max[|\min_i x_i|, |\max_i x_i|]\} < \theta $$ and this is zero if $\theta$ is too small, that is, lesser than $\max[|\min_i x_i|, |\max_i x_i|]$. At that point it becomes positive, and then decreasing from there. That gives the MLE as $$ \hat{\theta}_\text{MLE}=\max[|\min_i x_i|, |\max_i x_i|] $$ Another way to the solution is by noting that $|X_i| \sim \mathcal{U}(0,\theta)$ and use the solution for that case (which leads to an equivalent likelihood function), se for instance here

You must be more careful in deducing the likelihood function, otherwise your thinking seems correct. The density function for one $X_i$ is $$ f(x_i)=\frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} $$ So the likelihood function can be written as \begin{align*} \mathcal{L}(\theta)&=\prod_{i=1}^n \frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} \\ &=2^{-n}\theta^{-n} \cdot \prod_i I\{-\theta<X_i < \theta \} \\ &= 2^{-n}\theta^{-n} \cdot I\{-\theta < \min_i x_i \le \max_i x_i <\theta \} \\ &= 2^{-n}\theta^{-n} \cdot I\{0\le \max[|\min_i x_i|, |\max_i x_i|]\} < \theta \end{align*} and this is zero if $\theta$ is too small, that is, lesser than $\max[|\min_i x_i|, |\max_i x_i|]$. At that point it becomes positive, and then decreasing from there. That gives the MLE as $$ \hat{\theta}_\text{MLE}=\max[|\min_i x_i|, |\max_i x_i|] $$ Another way to the solution is by noting that $|X_i| \sim \mathcal{U}(0,\theta)$ and use the solution for that case (which leads to an equivalent likelihood function), se for instance here

You must be more careful in deducing the likelihood function, otherwise your thinking seems correct. The density function for one $X_i$ is $$ f(x_i)=\frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} $$ So the likelihood function can be written as $$ \mathcal{L}(\theta)=\prod_{i=1}^n \frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} \\ =2^{-n}\theta^{-n} \cdot \prod_i I\{-\theta<X_i < \theta \} \\ = 2^{-n}\theta^{-n} \cdot I\{-\theta < \min_i x_i \le \max_i x_i <\theta \} \\ = 2^{-n}\theta^{-n} \cdot I\{0\le \max[|\min_i x_i|, |\max_i x_i|]\} < \theta $$ and this is zero if $\theta$ is too small, that is, lesser than $\max[|\min_i x_i|, |\max_i x_i|]$. At that point it becomes positive, and then decreasing from there. That gives the MLE as $$ \hat{\theta}_\text{MLE}=\max[|\min_i x_i|, |\max_i x_i|] $$

You must be more careful in deducing the likelihood function, otherwise your thinking seems correct. The density function for one $X_i$ is $$ f(x_i)=\frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} $$ So the likelihood function can be written as $$ \mathcal{L}(\theta)=\prod_{i=1}^n \frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} \\ =2^{-n}\theta^{-n} \cdot \prod_i I\{-\theta<X_i < \theta \} \\ = 2^{-n}\theta^{-n} \cdot I\{-\theta < \min_i x_i \le \max_i x_i <\theta \} \\ = 2^{-n}\theta^{-n} \cdot I\{0\le \max[|\min_i x_i|, |\max_i x_i|]\} < \theta $$ and this is zero if $\theta$ is too small, that is, lesser than $\max[|\min_i x_i|, |\max_i x_i|]$. At that point it becomes positive, and then decreasing from there. That gives the MLE as $$ \hat{\theta}_\text{MLE}=\max[|\min_i x_i|, |\max_i x_i|] $$ Another way to the solution is by noting that $|X_i| \sim \mathcal{U}(0,\theta)$ and use the solution for that case (which leads to an equivalent likelihood function), se for instance here